# The speed of a projectile when it reaches its maximum height

## Homework Statement

The speed of a projectile when it reaches its maximum height is one-half its speed when it is at half its maxi- mum height. What is the initial projection angle of the projectile?

Vf2 = vi2 + 2ax

## The Attempt at a Solution

I know that the speed at the top only consist of an x component and that the speed at half the height would be sqrt(vx2+vy2) . But im trying everything from two hours and not getting a solution. The answer is suppost to be 67.8 but I dont know how to get it someone help

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The max height of the projectile is given by $\frac {u^2sin^2(\theta)}{2g}$ where $\theta$ is the angle of projection. Half the height would be $\frac {u^2sin^2(\theta)}{4g}$ . Find the y-component of velocity at half the height . Find the speed by adding the x and y components. Equate it to twice the x-component. Solve for $\theta$ . You should get your answer.

Are those the correct components?

The x-component is correct ($v_icos (\theta)$) the y-component is wrong. You should learn $\LaTeX$ else your equations are hard to read.

You are just giving me equations, I don't think you are putting in much effort. For the y-component $v_i = v_isin(\theta)$ , acceleration $a = -g$ and height $h = \frac{v_i^2sin^2(\theta)}{4g}$ What equation of motion do you use now to get $v_f$ ?

Vf2 = vi2 + 2ax

You have clearly made a calculation mistake , using the equation you'll get $v_f^2 = \frac{v_i^2sin^2(\theta)}{2}$ . The resulting equation will be $$4v_i^2cos^2(\theta) = v_i^2cos^2(\theta) + \frac{v_i^2sin^2(\theta)}{2}$$ Solve for $\theta$ you'll get your answer.